Geometry problem

Hello!
I can't seem to solve through this problem. I have a triangle (pic attached). I need to show EF is parallel to BC given BD=DC, and angle EDB = angle ADE, and angle ADF = angle FDC. How do I do it? I really tried for a while, and here are some of my observations:

Angle EDF is a right angle since 2*alpha+2*beta = 180, therefore alpha+beta = 90. And I'm trying to show equality of alternate angles but with no success.

First construct two rhombi, DGES and DHFT, as you see on the attached picture (construction is possible since DE bisects angle BDA and DF bisects angle CDA). Because they are rhombi, we have: GD=ES and they are parallel, DH=TF and they are parallel.
Because of some similar triangles and because of BD=DC, you can check the following calculation holds: AS/DS = AS/ES = AD/BD = AD/DC = AT/TF = AT/DT.
But this means that S=T and we are done, because we have: E,S=T,F lie all on one line which is parallel to BC.